At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 8 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate (in ft/min) is the height of the pile changing when the pile is 12 feet high? (Hint: The formula for the volume of a cone is V = 1 3 πr2h.)

dh/dt  =  0,008 ft/min

Step-by-step explanation:

The volume of the cone is:

V(c) =( 1/3)*π*r²*h     (1)

Where r is radius of the base

We know from problem statement

dV/dt = 8 ft³/min

And d =3*h    ⇒  2*r = 3*h     ⇒  r = (3/2)*h

Plugging the value of r in equation (1) we get

V = (1/3)*π*[ (3/2)*h ]²*h

V = (3/4)*π*h³

Now we differentiate relation to time , on both sides of the equation to get

dV/dt = (3/4)*π*3*h²*dh/dt

dV/dt = (9/4)*π*h²*dh/dt    

The question is dh/dt when h = 12 ft. Therefore

8  =  (9/4)* 3,14* (12)² *dh/dt    ⇒  dh/dt  =  8 / 1017,36

dh/dt  =  0,008 ft/min

b

step-by-step explanation:

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this